# Two schools A and B decided to award prices to their students for 3 values, honesty(X),punctuality(Y), and obedience(Z). School A decided to award a total of Rs.11000 for the three values to 5,4,and 3 students while school B decided to award Rs.10,700 for the 3 values to be 4,3,5 2students . If all the 3 prices together amount to be Rs. 2700, then: 1. Represent the abuve situation by a matric waequation and form linear equations using matrix multiplication.2. Is it possible to solve the system of equations so obtained using matrices?3. Which value do you nprefer to be rewarded and why?

The answer posted by @frienz-fun is correct except for the value 1100. It must be 11000.

@frienz-fun: Keep up the hard work!! Keep posting.!

• -4

school A  5x + 4y + 3z = 1100

school B  4x + 3y + 5z = 10700

all three prizes together cost  x + y + z = 2700

so in matrix form- AX = B

1.A is the matrix of order 3x3 containing coefficients of x,y,z of th eequations.

X is the matrix of order 3x1 containing x,y,z in a column.

B is the matrix of order 3x1 containing the costs 1100,10700,2700 in a column.

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2. from the matrix A check if it is invertible.(i.e., if det.A = 0)

if yes, then the system of equations can b solved using matrices.

• 3

preference of value can b accd to u....afterall all values are of same importance.

u cn give any one value provided u justify ur opinion.!!

hope it helps!!!!

• 8

THanx a lot .. :))

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